Matplotlib: Render Patch above Annotation - python

I have several patches (circles) which are labeled with a text as annotation inside. The two circles are draggable so it is possible, that these overlap. The problem is, that the all circles are always rendered before all annotations, so that all annotations are rendered above the circles. This means, if circle 1 overlaps circle 2, the annotations of both circles are visible. This looks kinda weird and wrong. I tried to solve this with the use of the zorder parameter, but this parameter had no effect. After further studying the documentation, this behaviour seems to be hardcoded into the renderer and I haven't found any solution to change this behaviour.
This is the current way how I draw a single circle with annotation:
def draw_circle_with_annotation(ax, position, label, point_size):
trans = ScaledTranslation(position[0], position[1], ax.transData)
point = Circle((0, 0), point_size, transform=trans, picker=True)
ax.add_patch(self.point)
label = ax.annotate(label, xy=(0.5, 0.5), xycoords=point, fontsize=10, ha='center', va='center_baseline')
This function is called successively for each circle, so that the draw calls of the patches and the annotations are executed alternately. Still all annotations are rendered on top of all patches.
However, is there any way to draw a patch above an annotation?

Matplotlib first draws elements with the lowest zorder, and then subsequently the higher zorders on top. When elements have the same zorder, patches are draw before texts. The solution is to give every annotation a new, higher zorder:
import matplotlib.pyplot as plt
from matplotlib.transforms import ScaledTranslation
from matplotlib.patches import Circle
import numpy as np
zorder = 2
def draw_circle_with_annotation(ax, position, label, point_size):
global zorder
point = Circle(position, point_size, picker=True, facecolor='tomato', edgecolor='k', zorder=zorder)
ax.add_patch(point)
ax.annotate(label, xy=position, fontsize=14, ha='center', va='center_baseline', zorder=zorder)
zorder += 1
fig, ax = plt.subplots()
for i in range(10):
draw_circle_with_annotation(ax, np.random.uniform(10, [50, 25], 2), f'label\n{i + 1}', 5)
ax.autoscale()
ax.set_aspect('equal')
plt.show()

Related

Error when trying to Move Figure Legend in MatplotLib [duplicate]

I have a series of 20 plots (not subplots) to be made in a single figure. I want the legend to be outside of the box. At the same time, I do not want to change the axes, as the size of the figure gets reduced.
I want to keep the legend box outside the plot area (I want the legend to be outside at the right side of the plot area).
Is there a way to reduce the font size of the text inside the legend box, so that the size of the legend box will be small?
There are a number of ways to do what you want. To add to what Christian Alis and Navi already said, you can use the bbox_to_anchor keyword argument to place the legend partially outside the axes and/or decrease the font size.
Before you consider decreasing the font size (which can make things awfully hard to read), try playing around with placing the legend in different places:
So, let's start with a generic example:
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
ax.plot(x, i * x, label='$y = %ix$' % i)
ax.legend()
plt.show()
If we do the same thing, but use the bbox_to_anchor keyword argument we can shift the legend slightly outside the axes boundaries:
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
ax.plot(x, i * x, label='$y = %ix$' % i)
ax.legend(bbox_to_anchor=(1.1, 1.05))
plt.show()
Similarly, make the legend more horizontal and/or put it at the top of the figure (I'm also turning on rounded corners and a simple drop shadow):
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
line, = ax.plot(x, i * x, label='$y = %ix$'%i)
ax.legend(loc='upper center', bbox_to_anchor=(0.5, 1.05),
ncol=3, fancybox=True, shadow=True)
plt.show()
Alternatively, shrink the current plot's width, and put the legend entirely outside the axis of the figure (note: if you use tight_layout(), then leave out ax.set_position():
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
ax.plot(x, i * x, label='$y = %ix$'%i)
# Shrink current axis by 20%
box = ax.get_position()
ax.set_position([box.x0, box.y0, box.width * 0.8, box.height])
# Put a legend to the right of the current axis
ax.legend(loc='center left', bbox_to_anchor=(1, 0.5))
plt.show()
And in a similar manner, shrink the plot vertically, and put a horizontal legend at the bottom:
import matplotlib.pyplot as plt
import numpy as np
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
line, = ax.plot(x, i * x, label='$y = %ix$'%i)
# Shrink current axis's height by 10% on the bottom
box = ax.get_position()
ax.set_position([box.x0, box.y0 + box.height * 0.1,
box.width, box.height * 0.9])
# Put a legend below current axis
ax.legend(loc='upper center', bbox_to_anchor=(0.5, -0.05),
fancybox=True, shadow=True, ncol=5)
plt.show()
Have a look at the matplotlib legend guide. You might also take a look at plt.figlegend().
Placing the legend (bbox_to_anchor)
A legend is positioned inside the bounding box of the axes using the loc argument to plt.legend.
E.g., loc="upper right" places the legend in the upper right corner of the bounding box, which by default extents from (0, 0) to (1, 1) in axes coordinates (or in bounding box notation (x0, y0, width, height) = (0, 0, 1, 1)).
To place the legend outside of the axes bounding box, one may specify a tuple (x0, y0) of axes coordinates of the lower left corner of the legend.
plt.legend(loc=(1.04, 0))
A more versatile approach is to manually specify the bounding box into which the legend should be placed, using the bbox_to_anchor argument. One can restrict oneself to supply only the (x0, y0) part of the bbox. This creates a zero span box, out of which the legend will expand in the direction given by the loc argument. E.g.,
plt.legend(bbox_to_anchor=(1.04, 1), loc="upper left")
places the legend outside the axes, such that the upper left corner of the legend is at position (1.04, 1) in axes coordinates.
Further examples are given below, where additionally the interplay between different arguments like mode and ncols are shown.
l1 = plt.legend(bbox_to_anchor=(1.04, 1), borderaxespad=0)
l2 = plt.legend(bbox_to_anchor=(1.04, 0), loc="lower left", borderaxespad=0)
l3 = plt.legend(bbox_to_anchor=(1.04, 0.5), loc="center left", borderaxespad=0)
l4 = plt.legend(bbox_to_anchor=(0, 1.02, 1, 0.2), loc="lower left",
mode="expand", borderaxespad=0, ncol=3)
l5 = plt.legend(bbox_to_anchor=(1, 0), loc="lower right",
bbox_transform=fig.transFigure, ncol=3)
l6 = plt.legend(bbox_to_anchor=(0.4, 0.8), loc="upper right")
Details about how to interpret the 4-tuple argument to bbox_to_anchor, as in l4, can be found in this question. The mode="expand" expands the legend horizontally inside the bounding box given by the 4-tuple. For a vertically expanded legend, see this question.
Sometimes it may be useful to specify the bounding box in figure coordinates instead of axes coordinates. This is shown in the example l5 from above, where the bbox_transform argument is used to put the legend in the lower left corner of the figure.
Postprocessing
Having placed the legend outside the axes often leads to the undesired situation that it is completely or partially outside the figure canvas.
Solutions to this problem are:
Adjust the subplot parameters
One can adjust the subplot parameters such, that the axes take less space inside the figure (and thereby leave more space to the legend) by using plt.subplots_adjust. E.g.,
plt.subplots_adjust(right=0.7)
leaves 30% space on the right-hand side of the figure, where one could place the legend.
Tight layout
Using plt.tight_layout Allows to automatically adjust the subplot parameters such that the elements in the figure sit tight against the figure edges. Unfortunately, the legend is not taken into account in this automatism, but we can supply a rectangle box that the whole subplots area (including labels) will fit into.
plt.tight_layout(rect=[0, 0, 0.75, 1])
Saving the figure with bbox_inches = "tight"
The argument bbox_inches = "tight" to plt.savefig can be used to save the figure such that all artist on the canvas (including the legend) are fit into the saved area. If needed, the figure size is automatically adjusted.
plt.savefig("output.png", bbox_inches="tight")
Automatically adjusting the subplot parameters
A way to automatically adjust the subplot position such that the legend fits inside the canvas without changing the figure size can be found in this answer: Creating figure with exact size and no padding (and legend outside the axes)
Comparison between the cases discussed above:
Alternatives
A figure legend
One may use a legend to the figure instead of the axes, matplotlib.figure.Figure.legend. This has become especially useful for Matplotlib version 2.1 or later, where no special arguments are needed
fig.legend(loc=7)
to create a legend for all artists in the different axes of the figure. The legend is placed using the loc argument, similar to how it is placed inside an axes, but in reference to the whole figure - hence it will be outside the axes somewhat automatically. What remains is to adjust the subplots such that there is no overlap between the legend and the axes. Here the point "Adjust the subplot parameters" from above will be helpful. An example:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0, 2*np.pi)
colors = ["#7aa0c4", "#ca82e1", "#8bcd50", "#e18882"]
fig, axes = plt.subplots(ncols=2)
for i in range(4):
axes[i//2].plot(x, np.sin(x+i), color=colors[i], label="y=sin(x + {})".format(i))
fig.legend(loc=7)
fig.tight_layout()
fig.subplots_adjust(right=0.75)
plt.show()
Legend inside dedicated subplot axes
An alternative to using bbox_to_anchor would be to place the legend in its dedicated subplot axes (lax).
Since the legend subplot should be smaller than the plot, we may use gridspec_kw={"width_ratios":[4, 1]} at axes creation.
We can hide the axes lax.axis("off"), but we still put a legend in. The legend handles and labels need to obtained from the real plot via h, l = ax.get_legend_handles_labels() and can then be supplied to the legend in the lax subplot, lax.legend(h, l). A complete example is below.
import matplotlib.pyplot as plt
plt.rcParams["figure.figsize"] = 6, 2
fig, (ax, lax) = plt.subplots(ncols=2, gridspec_kw={"width_ratios":[4, 1]})
ax.plot(x, y, label="y=sin(x)")
....
h, l = ax.get_legend_handles_labels()
lax.legend(h, l, borderaxespad=0)
lax.axis("off")
plt.tight_layout()
plt.show()
This produces a plot which is visually pretty similar to the plot from above:
We could also use the first axes to place the legend, but use the bbox_transform of the legend axes,
ax.legend(bbox_to_anchor=(0, 0, 1, 1), bbox_transform=lax.transAxes)
lax.axis("off")
In this approach, we do not need to obtain the legend handles externally, but we need to specify the bbox_to_anchor argument.
Further reading and notes:
Consider the Matplotlib legend guide with some examples of other stuff you want to do with legends.
Some example code for placing legends for pie charts may directly be found in answer to this question: Python - Legend overlaps with the pie chart
The loc argument can take numbers instead of strings, which make calls shorter, however, they are not very intuitively mapped to each other. Here is the mapping for reference:
Just call legend() after the plot() call like this:
# Matplotlib
plt.plot(...)
plt.legend(loc='center left', bbox_to_anchor=(1, 0.5))
# Pandas
df.myCol.plot().legend(loc='center left', bbox_to_anchor=(1, 0.5))
Results would look something like this:
You can make the legend text smaller by specifying set_size of FontProperties.
Resources:
Legend guide
matplotlib.legend
matplotlib.pyplot.legend
matplotlib.font_manager
set_size(self, size)
Valid font size are xx-small, x-small, small, medium, large, x-large, xx-large, larger, smaller, and None.
Real Python: Python Plotting With Matplotlib (Guide)
import matplotlib.pyplot as plt
from matplotlib.font_manager import FontProperties
fontP = FontProperties()
fontP.set_size('xx-small')
p1, = plt.plot([1, 2, 3], label='Line 1')
p2, = plt.plot([3, 2, 1], label='Line 2')
plt.legend(handles=[p1, p2], title='title', bbox_to_anchor=(1.05, 1), loc='upper left', prop=fontP)
fontsize='xx-small' also works, without importing FontProperties.
plt.legend(handles=[p1, p2], title='title', bbox_to_anchor=(1.05, 1), loc='upper left', fontsize='xx-small')
To place the legend outside the plot area, use loc and bbox_to_anchor keywords of legend(). For example, the following code will place the legend to the right of the plot area:
legend(loc="upper left", bbox_to_anchor=(1,1))
For more info, see the legend guide
Short answer: you can use bbox_to_anchor + bbox_extra_artists + bbox_inches='tight'.
Longer answer:
You can use bbox_to_anchor to manually specify the location of the legend box, as some other people have pointed out in the answers.
However, the usual issue is that the legend box is cropped, e.g.:
import matplotlib.pyplot as plt
# data
all_x = [10,20,30]
all_y = [[1,3], [1.5,2.9],[3,2]]
# Plot
fig = plt.figure(1)
ax = fig.add_subplot(111)
ax.plot(all_x, all_y)
# Add legend, title and axis labels
lgd = ax.legend( [ 'Lag ' + str(lag) for lag in all_x], loc='center right', bbox_to_anchor=(1.3, 0.5))
ax.set_title('Title')
ax.set_xlabel('x label')
ax.set_ylabel('y label')
fig.savefig('image_output.png', dpi=300, format='png')
In order to prevent the legend box from getting cropped, when you save the figure you can use the parameters bbox_extra_artists and bbox_inches to ask savefig to include cropped elements in the saved image:
fig.savefig('image_output.png', bbox_extra_artists=(lgd,), bbox_inches='tight')
Example (I only changed the last line to add 2 parameters to fig.savefig()):
import matplotlib.pyplot as plt
# data
all_x = [10,20,30]
all_y = [[1,3], [1.5,2.9],[3,2]]
# Plot
fig = plt.figure(1)
ax = fig.add_subplot(111)
ax.plot(all_x, all_y)
# Add legend, title and axis labels
lgd = ax.legend( [ 'Lag ' + str(lag) for lag in all_x], loc='center right', bbox_to_anchor=(1.3, 0.5))
ax.set_title('Title')
ax.set_xlabel('x label')
ax.set_ylabel('y label')
fig.savefig('image_output.png', dpi=300, format='png', bbox_extra_artists=(lgd,), bbox_inches='tight')
I wish that matplotlib would natively allow outside location for the legend box as Matlab does:
figure
x = 0:.2:12;
plot(x,besselj(1,x),x,besselj(2,x),x,besselj(3,x));
hleg = legend('First','Second','Third',...
'Location','NorthEastOutside')
% Make the text of the legend italic and color it brown
set(hleg,'FontAngle','italic','TextColor',[.3,.2,.1])
In addition to all the excellent answers here, newer versions of matplotlib and pylab can automatically determine where to put the legend without interfering with the plots, if possible.
pylab.legend(loc='best')
This will automatically place the legend away from the data if possible!
However, if there isn't any place to put the legend without overlapping the data, then you'll want to try one of the other answers; using loc="best" will never put the legend outside of the plot.
Short Answer: Invoke draggable on the legend and interactively move it wherever you want:
ax.legend().draggable()
Long Answer: If you rather prefer to place the legend interactively/manually rather than programmatically, you can toggle the draggable mode of the legend so that you can drag it to wherever you want. Check the example below:
import matplotlib.pylab as plt
import numpy as np
#define the figure and get an axes instance
fig = plt.figure()
ax = fig.add_subplot(111)
#plot the data
x = np.arange(-5, 6)
ax.plot(x, x*x, label='y = x^2')
ax.plot(x, x*x*x, label='y = x^3')
ax.legend().draggable()
plt.show()
Newer versions of Matplotlib have made it much easier to position the legend outside the plot. I produced this example with Matplotlib version 3.1.1.
Users can pass a 2-tuple of coordinates to the loc parameter to position the legend anywhere in the bounding box. The only gotcha is you need to run plt.tight_layout() to get matplotlib to recompute the plot dimensions so the legend is visible:
import matplotlib.pyplot as plt
plt.plot([0, 1], [0, 1], label="Label 1")
plt.plot([0, 1], [0, 2], label='Label 2')
plt.legend(loc=(1.05, 0.5))
plt.tight_layout()
This leads to the following plot:
References:
matplotlib.pyplot.legend
It is not exactly what you asked for, but I found it's an alternative for the same problem.
Make the legend semitransparent, like so:
Do this with:
fig = pylab.figure()
ax = fig.add_subplot(111)
ax.plot(x, y, label=label, color=color)
# Make the legend transparent:
ax.legend(loc=2, fontsize=10, fancybox=True).get_frame().set_alpha(0.5)
# Make a transparent text box
ax.text(0.02, 0.02, yourstring, verticalalignment='bottom',
horizontalalignment='left',
fontsize=10,
bbox={'facecolor':'white', 'alpha':0.6, 'pad':10},
transform=self.ax.transAxes)
As noted, you could also place the legend in the plot, or slightly off it to the edge as well. Here is an example using the Plotly Python API, made with an IPython Notebook. I'm on the team.
To begin, you'll want to install the necessary packages:
import plotly
import math
import random
import numpy as np
Then, install Plotly:
un='IPython.Demo'
k='1fw3zw2o13'
py = plotly.plotly(username=un, key=k)
def sin(x,n):
sine = 0
for i in range(n):
sign = (-1)**i
sine = sine + ((x**(2.0*i+1))/math.factorial(2*i+1))*sign
return sine
x = np.arange(-12,12,0.1)
anno = {
'text': '$\\sum_{k=0}^{\\infty} \\frac {(-1)^k x^{1+2k}}{(1 + 2k)!}$',
'x': 0.3, 'y': 0.6,'xref': "paper", 'yref': "paper",'showarrow': False,
'font':{'size':24}
}
l = {
'annotations': [anno],
'title': 'Taylor series of sine',
'xaxis':{'ticks':'','linecolor':'white','showgrid':False,'zeroline':False},
'yaxis':{'ticks':'','linecolor':'white','showgrid':False,'zeroline':False},
'legend':{'font':{'size':16},'bordercolor':'white','bgcolor':'#fcfcfc'}
}
py.iplot([{'x':x, 'y':sin(x,1), 'line':{'color':'#e377c2'}, 'name':'$x\\\\$'},\
{'x':x, 'y':sin(x,2), 'line':{'color':'#7f7f7f'},'name':'$ x-\\frac{x^3}{6}$'},\
{'x':x, 'y':sin(x,3), 'line':{'color':'#bcbd22'},'name':'$ x-\\frac{x^3}{6}+\\frac{x^5}{120}$'},\
{'x':x, 'y':sin(x,4), 'line':{'color':'#17becf'},'name':'$ x-\\frac{x^5}{120}$'}], layout=l)
This creates your graph, and allows you a chance to keep the legend within the plot itself. The default for the legend if it is not set is to place it in the plot, as shown here.
For an alternative placement, you can closely align the edge of the graph and border of the legend, and remove border lines for a closer fit.
You can move and re-style the legend and graph with code, or with the GUI. To shift the legend, you have the following options to position the legend inside the graph by assigning x and y values of <= 1. E.g :
{"x" : 0,"y" : 0} -- Bottom Left
{"x" : 1, "y" : 0} -- Bottom Right
{"x" : 1, "y" : 1} -- Top Right
{"x" : 0, "y" : 1} -- Top Left
{"x" :.5, "y" : 0} -- Bottom Center
{"x": .5, "y" : 1} -- Top Center
In this case, we choose the upper right, legendstyle = {"x" : 1, "y" : 1}, also described in the documentation:
I simply used the string 'center left' for the location, like in MATLAB.
I imported pylab from Matplotlib.
See the code as follows:
from matplotlib as plt
from matplotlib.font_manager import FontProperties
t = A[:, 0]
sensors = A[:, index_lst]
for i in range(sensors.shape[1]):
plt.plot(t, sensors[:, i])
plt.xlabel('s')
plt.ylabel('°C')
lgd = plt.legend(loc='center left', bbox_to_anchor=(1, 0.5), fancybox = True, shadow = True)
You can also try figlegend. It is possible to create a legend independent of any Axes object. However, you may need to create some "dummy" Paths to make sure the formatting for the objects gets passed on correctly.
Something along these lines worked for me. Starting with a bit of code taken from Joe, this method modifies the window width to automatically fit a legend to the right of the figure.
import matplotlib.pyplot as plt
import numpy as np
plt.ion()
x = np.arange(10)
fig = plt.figure()
ax = plt.subplot(111)
for i in xrange(5):
ax.plot(x, i * x, label='$y = %ix$'%i)
# Put a legend to the right of the current axis
leg = ax.legend(loc='center left', bbox_to_anchor=(1, 0.5))
plt.draw()
# Get the ax dimensions.
box = ax.get_position()
xlocs = (box.x0,box.x1)
ylocs = (box.y0,box.y1)
# Get the figure size in inches and the dpi.
w, h = fig.get_size_inches()
dpi = fig.get_dpi()
# Get the legend size, calculate new window width and change the figure size.
legWidth = leg.get_window_extent().width
winWidthNew = w*dpi+legWidth
fig.set_size_inches(winWidthNew/dpi,h)
# Adjust the window size to fit the figure.
mgr = plt.get_current_fig_manager()
mgr.window.wm_geometry("%ix%i"%(winWidthNew,mgr.window.winfo_height()))
# Rescale the ax to keep its original size.
factor = w*dpi/winWidthNew
x0 = xlocs[0]*factor
x1 = xlocs[1]*factor
width = box.width*factor
ax.set_position([x0,ylocs[0],x1-x0,ylocs[1]-ylocs[0]])
plt.draw()
New in matplotlib 3.7
Legends now accept "outside" locations directly, e.g., loc='outside right upper'.
Just make sure the layout is constrained and then prepend "outside" to the loc string:
import matplotlib.pyplot as plt
import numpy as np
fig, ax = plt.subplots(layout='constrained')
# --------------------
x = np.linspace(-np.pi, np.pi)
ax.plot(x, x, label='$f(x) = x$')
ax.plot(x, np.sin(x), label='$f(x) = sin(x)$')
ax.plot(x, np.cos(x), label='$f(x) = cos(x)$')
fig.legend(loc='outside right upper')
# -------
plt.show()
Multiple subplots also work fine with the new "outside" locations:
fig, (ax1, ax2) = plt.subplots(1, 2, layout='constrained')
# --------------------
x = np.linspace(-np.pi, np.pi)
ax1.plot(x, x, '-', label='$f(x) = x$')
ax1.plot(x, np.sin(x), '--', label='$f(x) = sin(x)$')
ax2.plot(x, np.cos(x), ':', label='$f(x) = cos(x)$')
fig.legend(loc='outside right center')
# -------
Of course the available "outside" locations are preset, so use the older answers if you need finer positioning. However the standard locations should fit most use cases:
locs = [
'outside upper left', 'outside upper center', 'outside upper right',
'outside center right', 'upper center left',
'outside lower right', 'outside lower center', 'outside lower left',
]
for loc in locs:
fig.legend(loc=loc, title=loc)
locs = [
'outside right upper', 'outside right lower',
'outside left lower', 'outside left upper',
]
for loc in locs:
fig.legend(loc=loc, title=loc)
The solution that worked for me when I had a huge legend was to use an extra empty image layout.
In the following example, I made four rows and at the bottom I plotted the image with an offset for the legend (bbox_to_anchor). At the top it does not get cut.
f = plt.figure()
ax = f.add_subplot(414)
lgd = ax.legend(loc='upper left', bbox_to_anchor=(0, 4), mode="expand", borderaxespad=0.3)
ax.autoscale_view()
plt.savefig(fig_name, format='svg', dpi=1200, bbox_extra_artists=(lgd,), bbox_inches='tight')
Here's another solution, similar to adding bbox_extra_artists and bbox_inches, where you don't have to have your extra artists in the scope of your savefig call. I came up with this since I generate most of my plot inside functions.
Instead of adding all your additions to the bounding box when you want to write it out, you can add them ahead of time to the Figure's artists. Using something similar to Franck Dernoncourt's answer:
import matplotlib.pyplot as plt
# Data
all_x = [10, 20, 30]
all_y = [[1, 3], [1.5, 2.9], [3, 2]]
# Plotting function
def gen_plot(x, y):
fig = plt.figure(1)
ax = fig.add_subplot(111)
ax.plot(all_x, all_y)
lgd = ax.legend(["Lag " + str(lag) for lag in all_x], loc="center right", bbox_to_anchor=(1.3, 0.5))
fig.artists.append(lgd) # Here's the change
ax.set_title("Title")
ax.set_xlabel("x label")
ax.set_ylabel("y label")
return fig
# Plotting
fig = gen_plot(all_x, all_y)
# No need for `bbox_extra_artists`
fig.savefig("image_output.png", dpi=300, format="png", bbox_inches="tight")
.
Here is an example from the matplotlib tutorial found here. This is one of the more simpler examples but I added transparency to the legend and added plt.show() so you can paste this into the interactive shell and get a result:
import matplotlib.pyplot as plt
p1, = plt.plot([1, 2, 3])
p2, = plt.plot([3, 2, 1])
p3, = plt.plot([2, 3, 1])
plt.legend([p2, p1, p3], ["line 1", "line 2", "line 3"]).get_frame().set_alpha(0.5)
plt.show()

is it possible to draw an arrow from somewhere below ticks in matplotlib

I am trying to draw a plot like this.
the tricky thing is the arrow from "z = -1.48" to x axis, which is under the ticks.
the piece of code shown below is my trying to draw that arrow.
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
fig, ax = plt.subplots(figsize=(8,3))
x = np.linspace(-5,5,101)
x2 = np.linspace(-5,-1.5,101)
y = norm.pdf(x)
y2 = norm.pdf(x2)
t = ax.text(-3, 0.15, '-1.5')
plt.grid(True)
ax.plot(x, norm.pdf(x))
ax.fill_between(x2, 0, y2, color='r')
unfortunately, the part under ticks doesn't render.
is it possible to make the arrow under ticks shown?
You want annotate:
ax.annotate('1.5',
xy=(-1.5, 0),
xytext=(-1.5, -0.1),
ha='center',
va='center',
arrowprops={'arrowstyle': '->'})
This places text at xytext and draws an arrow from there to xy. ha and va refer to the alignment of the text based on those coordinates. If you don't pass 'center', the arrow and the text will be slightly misaligned. Lastly, passing arrowprops allows you to benefit from an inbuilt arrow style.
More information can be found in the docs.

Python matplotlib - add borders to grid plot based on value

I wonder if you can help me with this. I have a grid of 0's and 1's that I want to add a border colour to the plot cell area if it's a 1. I've used imshow to produce a grid coloured according to value, e.g.:
a = np.random.randint(2, size=(10,10))
im = plt.imshow(a, cmap='Blues', interpolation='none', vmin=0, vmax=1, aspect='equal')
However, I can't find any border properties to change for each grid cell in imshow. I've read add_patch could be used to place a rectangle at certain points to mimic a border using on the axes values, but is there a better way than looping and applying cell-wise?
Thanks for any help you can give.
There is no build-in function to partially colorize imshow edges. The easiest option is probably indeed to draw a rectangle for each cell.
import matplotlib.pyplot as plt
import numpy as np
a = np.random.randint(2, size=(10,10))
im = plt.imshow(a, cmap='Blues', interpolation='none', vmin=0, vmax=1, aspect='equal')
def rect(pos):
r = plt.Rectangle(pos-0.5, 1,1, facecolor="none", edgecolor="k", linewidth=2)
plt.gca().add_patch(r)
x,y = np.meshgrid(np.arange(a.shape[1]),np.arange(a.shape[0]))
m = np.c_[x[a.astype(bool)],y[a.astype(bool)]]
for pos in m:
rect(pos)
plt.show()

matplotlib - autosize of text according to shape size

I'm adding a text inside a shape by:
ax.text(x,y,'text', ha='center', va='center',bbox=dict(boxstyle='circle', fc="w", ec="k"),fontsize=10) (ax is AxesSubplot)
The problem is that I couldn't make the circle size constant while changing the string length. I want the text size adjust to the circle size and not the other way around.
The circle is even completely gone if the string is an empty one.
The only bypass to the problem I had found is dynamically to set the fontsize param according to the len of the string, but that's too ugly and not still the circle size is not completely constant.
EDIT (adding a MVCE):
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_axes([0,0,1,1])
ax.text(0.5,0.5,'long_text', ha='center', va='center',bbox=dict(boxstyle='circle', fc="w", ec="k"),fontsize=10)
ax.text(0.3,0.7,'short', ha='center', va='center',bbox=dict(boxstyle='circle', fc="w", ec="k"),fontsize=10)
plt.show()
Trying to make both circles the same size although the string len is different. Currently looks like this:
I have a very dirty and hard-core solution which requires quite deep knowledge of matplotlib. It is not perfect but might give you some ideas how to start.
import matplotlib.pyplot as plt
from matplotlib.patches import Circle
import numpy as np
plt.close('all')
fig, ax = plt.subplots(1, 1, figsize=(8, 8))
t1 = ax.text(0.5,0.5,'long_text', ha='center', va='center',fontsize=10)
t2 = ax.text(0.3,0.7,'short', ha='center', va='center', fontsize=10)
t3 = ax.text(0.1,0.7,'super-long-text-that-is-long', ha='center', va='center', fontsize=10)
fig.show()
def text_with_circle(text_obj, axis, color, border=1.5):
# Get the box containing the text
box1 = text_obj.get_window_extent()
# It turned out that what you get from the box is
# in screen pixels, so we need to transform them
# to "data"-coordinates. This is done with the
# transformer-function below
transformer = axis.transData.inverted().transform
# Now transform the corner coordinates of the box
# to data-coordinates
[x0, y0] = transformer([box1.x0, box1.y0])
[x1, y1] = transformer([box1.x1, box1.y1])
# Find the x and y center coordinate
x_center = (x0+x1)/2.
y_center = (y0+y1)/2.
# Find the radius, add some extra to make a nice border around it
r = np.max([(x1-x0)/2., (y1-y0)/2.])*border
# Plot the a circle at the center of the text, with radius r.
circle = Circle((x_center, y_center), r, color=color)
# Add the circle to the axis.
# Redraw the canvas.
return circle
circle1 = text_with_circle(t1, ax, 'g')
ax.add_artist(circle1)
circle2 = text_with_circle(t2, ax, 'r', 5)
ax.add_artist(circle2)
circle3 = text_with_circle(t3, ax, 'y', 1.1)
ax.add_artist(circle3)
fig.canvas.draw()
At the moment you have to run this in ipython, because the figure has to be drawn BEFORE you get_window_extent(). Therefore the fig.show() has to be called AFTER the text is added, but BEFORE the circle can be drawn! Then we can get the coordinates of the text, figures out where the middle is and add a circle around the text with a certain radius. When this is done we redraw the canvas to update with the new circle. Ofcourse you can customize the circle a lot more (edge color, face color, line width, etc), look into the Circle class.
Example of output plot:

Plotting a circle-matplotlib

I am trying to plot a circle on a grid. The code that I have written is as follows:
import pyplot as plt
from pyplot import Figure, subplot
fig=plt.figure(1)
plt.axis([0,400,0,400])
ax=fig.add_subplot(1,1,1)
circ=plt.Circle((200,200), radius=10, color='g', fill=False)
ax.add_patch(circ)
plt.show()
Now, I want the center of the circle to be the center of the graph, that is, (200,200) in this example. In case of other cases I want it to automatically choose the centre depending on the size that us set. Can this be in some way?
To make it clearer I want to get the x-axis and the y-axis range so as to find the mid point of the grid. How do I proceed?
Your x-axis and y-axis ranges are in your code right here:
plt.axis([0,400,0,400])
So all you would need is leverage on this a bit like so:
x_min = 0
x_max = 400
y_min = 0
y_max = 400
circle_x = (x_max-x_min)/2.
circle_y = (y_max-y_min)/2.
circ=plt.Circle((circle_x,circle_y), radius=10, color='g', fill=False)
If you want to catch x_min etc. from the command prompt then read out sys.argv.
What you want may be ax.transAxes, here's the tutorial for coordinates transformation.
ax.transAxes means the coordinate system of the Axes; (0,0) is bottom left of the axes, and (1,1) is top right of the axes.
fig=plt.figure(1)
plt.axis([0,400,0,400])
ax=fig.add_subplot(1,1,1)
circ=plt.Circle((0.5,0.5), radius=0.2, color='g', fill=False,transform=ax.transAxes)
ax.add_patch(circ)
plt.show()
Note that the radius is also transformed into Axes coordinate. If you specify a radius larger than sqrt(2)/2 (about 0.7) you will see nothing in the plot.
If you want to plot a set of circles, it would be much simpler if you use the function circles here. For this problem,
fig=plt.figure(1)
plt.axis([0,400,0,400])
ax=fig.add_subplot(1,1,1)
circles(0.5, 0.5, 0.2, c='g', ax=ax, facecolor='none', transform=ax.transAxes)
plt.show()
A bit more, if you want see a real circle (instead of an ellipse) in your figure, you should use
ax=fig.add_subplot(1,1,1, aspect='equal')
or something like that.

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